Field extension and algebraic element

Question

Suppose $K/F$ is a field extension and $a∈K$. If $a∈F(a^n)$ for some $n>1$, prove that $a$ is algebraic over $F$.

I know that if $a\in F(a^n)$, then $\{a^m : ∀m \in \mathbb{N}\}⊂F(a^n)$, so I can let $b∈F(a^n)$ with $ab=1$. If I can tell $b$ is the linear combination of $\{a^m:∀m∈\mathbb{N}\}$, then $a$ is algebraic on F. but I can't work it out in this way.

Answer 1

There are nonzero relatively prime polynomials $f(X),g(X)\in F[X]$ (where $X$ is an indeterminate) such that $$ a=\frac{f(a^n)}{g(a^n)}\quad. $$ Setting $h(X):=Xg(X^n)-f(X^n)\in F[X]$, we get $h(a)=0$.

It suffices to prove $h(X)\neq0$.

Assume by contradiction $h(X)=0$, that is $Xg(X^n)=f(X^n)$. This implies successively that $X$ divides $f(X)$, that $X^{n-1}$ divides $g(X^n)$, and that $X$ divides $g(X)$ (because $n\ge2$), contradicting the assumption that $f(X)$ and $g(X)$ are relatively prime.

Answer 2

$a\in F(a^n)$ implies $a={{P(a^n)}\over{Q(a^n)}}$ where $P$ and $Q$ are polynomials with coefficients in $F$. This implies that $P(a^n)-aQ(a^n)=0$ is an algebraic equation of $a$.